Function satisfying $\lim_{x\to 0}\frac{f(ax)}{f(x)}=a$
Solution 1:
I try a counter-example:
Let first be $g(x)=\sqrt{|\log (|x|)|}$. For any $a$ not zero $g(ax)-g(x)\to 0$ if $x\to 0$, $x\not =0$. (Because for $|x|$ small, $|\log |a|+\log |x||=-\log |x|-\log |a|$). Hence if we put $f(x)=x\exp(g(x))$ for $x\not =0$, and $f(0)=0$, we are done.
Solution 2:
However the previous answer is correct this gives a very simple one if $a $ may only be positive, then $f=|x|$ is also continuous and the limit of the ratio is also $a $ and the derivative doesn't exist in 0